Question: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $18$ years; the standard deviation is $1.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living less than $19.7$ years.
Solution: $18$ $16.3$ $19.7$ $14.6$ $21.4$ $12.9$ $23.1$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $18$ years. We know the standard deviation is $1.7$ years, so one standard deviation below the mean is $16.3$ years and one standard deviation above the mean is $19.7$ years. Two standard deviations below the mean is $14.6$ years and two standard deviations above the mean is $21.4$ years. Three standard deviations below the mean is $12.9$ years and three standard deviations above the mean is $23.1$ years. We are interested in the probability of a porcupine living less than $19.7$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the porcupines will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $16.3$ years and the other half $({16\%})$ will live longer than $19.7$ years. The probability of a particular porcupine living less than $19.7$ years is ${68\%} + {16\%}$, or $84\%$.